Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The following data are for the system $$ \mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ $$ \begin{array}{lcccccc} \hline \text { Time (s) } & 0 & 30 & 45 & 60 & 75 & 90 \\ P_{\mathrm{A}} \text { (atm) } & 0.500 & 0.390 & 0.360 & 0.340 & 0.325 & 0.325 \\\ P_{\text {B }} \text { (atm) } & 0.000 & 0.220 & 0.280 & 0.320 & 0.350 & 0.350 \\\ \hline \end{array} $$ (a) How long does it take the system to reach equilibrium? (b) How does the rate of the forward reaction compare with the rate of the reverse reaction after \(45 \mathrm{~s}\) ? After \(90 \mathrm{~s}\) ?

Short Answer

Expert verified
The system reaches equilibrium after 90 seconds. (b) Compare the forward and reverse reaction rates at 45s and 90s. At 45 seconds, the forward reaction rate is about 4.59 times greater than the reverse reaction rate. At 90 seconds (equilibrium), the forward reaction rate is equal to the reverse reaction rate.

Step by step solution

01

(a) Determine the time when the system reaches equilibrium

To determine when the system reaches equilibrium, we need to look for the time when the pressure of A and B no longer change and remain constant. By analyzing the pressure table, we can see that at time \textit{t=90 s}, the pressures of A and B are remain unchanged. Therefore, the equilibrium is reached after 90 seconds.
02

(b) Compare the forward and reverse reaction rates at 45s

Since the reaction is given by A ⇌ 2 B, the forward reaction rate will be proportional to the pressure of A (\(P_A\)), and the reverse reaction rate will be proportional to the square of the pressure of B (\(P_B^2\)) because there are two moles of B in the reaction equation. At 45 seconds, the pressures are as follows; \(P_A = 0.360~\text{atm}\) and \(P_B = 0.280~\text{atm}\) Calculate the ratio of the forward and reverse reaction rates at 45s: \(Ratio_{45s} = \frac{Rate_{forward}}{Rate_{reverse}} = \frac{P_A}{P_B^2} = \frac{0.360}{(0.280)^2} = 4.59\) So, the forward reaction rate is about 4.59 times greater than the reverse reaction rate at 45 seconds.
03

(b) Compare the forward and reverse reaction rates at 90s

At 90 seconds (equilibrium), the pressures are as follows; \(P_A = 0.325~\text{atm}\) and \(P_B = 0.350~\text{atm}\) Calculate the ratio of the forward and reverse reaction rates at 90s: \(Ratio_{90s} = \frac{Rate_{forward}}{Rate_{reverse}} = \frac{P_A}{P_B^2} = \frac{0.325}{(0.350)^2} = 2.65\) Since, at equilibrium, the forward reaction rate equals the reverse reaction rate, it means that our calculated ratio will be 1. This means that the forward reaction rate is equal to the reverse reaction rate at 90 seconds (they are the same).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ \(K\) at a certain temperature is \(3.7 \times 10^{-4}\). Predict the direction in which the system will move to reach equilibrium if one starts with $$ \begin{array}{l} \text { (a) } P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=P_{\mathrm{NH}_{3}}=0.01 \mathrm{~atm} \\ \text { (b) } P_{\mathrm{NH}_{3}}=0.0045 \mathrm{~atm} \end{array} $$ (c) \(P_{\mathrm{N}_{2}}=1.2 \mathrm{~atm}, P_{\mathrm{H}_{2}}=1.88 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=0.0058 \mathrm{~atm}\)

At a certain temperature, \(K=0.29\) for the decomposition of two moles of iodine trichloride, \(\mathrm{ICl}_{3}(s),\) to chlorine and iodine gases. The partial pressure of chlorine gas at equilibrium is three times that of iodine gas. What are the partial pressures of iodine and chlorine at equilibrium?

Write a chemical equation for an equilibrium system that would lead to the following expressions \((\mathrm{a}-\mathrm{d})\) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2} \mathrm{~S}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{2}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{Cl}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ When equilibrium is established, the partial pressures of the gases $$ \text { are: } P_{\mathrm{N}_{2}}=1.200 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.800 \mathrm{~atm}, P_{\mathrm{NO}}=0.0220 \mathrm{~atm} . $$ (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure 1.200 atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

At \(800^{\circ} \mathrm{C}, K=2.2 \times 10^{-4}\) for the following reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Calculate \(K\) at \(800^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) from \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) gases. (b) the decomposition of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) gas.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free